(4)int_((pi)/(4))^(pi)/(3)(x)/(sin^2)xdx;

为了求解定积分 $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x}{\sin^2 x} \, dx$，我们可以使用分部积分法。分部积分法的公式是： \[ \int u \, dv = uv - \int v \, du \] 在这个问题中，我们可以选择 $u = x$ 和 $dv = \frac{1}{\sin^2 x} \, dx = \csc^2 x \, dx$。那么，$du = dx$ 和 $v = -\cot x$（因为 $\int \csc^2 x \, dx = -\cot x$）。 应用分部积分法，我们得到： \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x}{\sin^2 x} \, dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} x \csc^2 x \, dx = \left[ x (-\cot x) \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} - \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (-\cot x) \, dx \] 简化后，我们有： \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x}{\sin^2 x} \, dx = \left[ -x \cot x \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot x \, dx \] 首先，我们计算 $\left[ -x \cot x \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$： \[ \left[ -x \cot x \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = -\frac{\pi}{3} \cot \frac{\pi}{3} - \left( -\frac{\pi}{4} \cot \frac{\pi}{4} \right) = -\frac{\pi}{3} \cdot \frac{1}{\sqrt{3}} + \frac{\pi}{4} \cdot 1 = -\frac{\pi}{3\sqrt{3}} + \frac{\pi}{4} = -\frac{\pi \sqrt{3}}{9} + \frac{\pi}{4} \] 接下来，我们计算 $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot x \, dx$： \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot x \, dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\cos x}{\sin x} \, dx \] 令 $u = \sin x$，则 $du = \cos x \, dx$。当 $x = \frac{\pi}{4}$ 时，$u = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$；当 $x = \frac{\pi}{3}$ 时，$u = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$。因此，积分变为： \[ \int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}} \frac{1}{u} \, du = \left[ \ln |u| \right]_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}} = \ln \frac{\sqrt{3}}{2} - \ln \frac{\sqrt{2}}{2} = \ln \left( \frac{\sqrt{3}}{2} \cdot \frac{2}{\sqrt{2}} \right) = \ln \sqrt{\frac{3}{2}} = \ln \left( \frac{\sqrt{6}}{2} \right) = \frac{1}{2} \ln \frac{3}{2} \] 将两部分结果相加，我们得到： \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x}{\sin^2 x} \, dx = -\frac{\pi \sqrt{3}}{9} + \frac{\pi}{4} + \frac{1}{2} \ln \frac{3}{2} \] 因此，最终答案是： \[ \boxed{\frac{\pi}{4} - \frac{\pi \sqrt{3}}{9} + \frac{1}{2} \ln \frac{3}{2}} \]